Heat Transfer Mechanisms — Question 6

Question 6

Net radiation subtracts the absorbed surroundings contribution

Use $\sigma=5.67\times10^{-8}\,\mathrm{W/(m^2K^4)}$. A gray surface has area $.100\,\mathrm{m^2}$, emissivity .800 and temperature 400 K. It faces a large black enclosure at 300 K, with unit view factor. Use $P_{\mathrm{net}}=\epsilon\sigma A(T^4-T_s^4)$; assume equal absorptivity and emissivity for the enclosure radiation.

1. Find emitted, absorbed and net outward powers.

2. If surface temperature is raised to 800 K while surroundings remain at 300 K, find the ratio of new to old net radiation. Why is it not exactly 16?

Question 6 - Solution

Net radiation subtracts the absorbed surroundings contribution

(a) Method and calculation

$$P_{\mathrm{emit}}=.8\sigma(.1)(400)^4=116.1216\,\mathrm W.$$$$P_{\mathrm{absorb}}=.8\sigma(.1)(300)^4=36.7416\,\mathrm W.$$$$P_{\mathrm{net}}=116.1216-36.7416=79.3800\,\mathrm W.$$

(b) Interpretation and further calculation

$$P_{\mathrm{net,800}}=.8\sigma(.1)(800^4-300^4)=1821.204\,\mathrm W.$$$$\frac{P_{\mathrm{net,800}}}{P_{\mathrm{net,400}}}=\frac{800^4-300^4}{400^4-300^4}=22.942857.$$Emission alone increases sixteenfold, but the same absorbed background is subtracted in both cases. All fourth powers require absolute temperature.

Physical check

At $T=T_s$, emission is nonzero but net radiation is zero.